6 – Electrochemistry
1] A spontaneous electrochemical cell would have which of the following? 1
a) Ecell = 0
b) Ecell > 0
c) Ecell < 0
d) Ecell  0
e) Ecell  0
2] The purpose of a reference electrode is to _________ 2
3] Which of the following species is the strongest reducing agent? 3
A + e- = AA- + e- = A2A2- + e- = A3-
E0 = 0.500 Volts
E0 = 0.000 volts
E0 = -0.500 volts
4] What is E0 for the following half reaction if E0 for 4
Zn2+ + 2e- = Zn(s)
is -0.762 V?
½ Zn2+ + e- = ½ Zn(s)
5] What is E0cell for the following reaction? 5
2Na(s) + 2H+ = 2Na+ + H2(g)
Na+ + e- = Na(s)
E0 = - 2.7143 V
2H+ + 2e- = H2(g)
E0 = 0.0000 V
6] If A + e- = B has E0 = 0.775 V then the E0 for 2A + 2e- = 2B is __________________. 6
7] The reductions take place at which electrode? 7
8] The standard cell potential for the following is 8
Fe(s)/Fe2+(aq)//Sn2+(aq)/Sn(s)
Fe2+ + 2e- = Fe(s)
E0 = -0.44 V
Sn2+ + 2e- = Sn(s)
E0 = -0.141 V
9] The E0 for the following is 9
FeCO3(s) + 2e- = Fe(s) + CO32Fe2+ + 2e- = Fe(s)
Ksp {FeCO3(s)} = 2.1e-11
E0 = ?
E0 = -0.44 V
10] What is E0cell for the reaction below 10
F2 + 2Fe2+ = 2F- + 2Fe3+
F2 + 2e- = 2FE0red = 2.890 V
Fe3+ + e- = Fe2+
E0red = 0.771 V
11] What is E0cell for the reaction below? 11
Hg2SO4(s) + 2e- = 2Hg(l) + SO42Hg22+ + 2e- = 2Hg(l)
E0red = 0.796 V
Hg2SO4(s) = Hg22+ + SO42Ksp = 7.4e-7
12] What is the half reaction reduction potential for 1.00e-5 M H+? 12
13] Which of the following species is the strongest reducing agent? 13
A+ + e- = A
E0 = 0.75 V
B + e- = BE0 = 0.25 V
D2+ + e- = D+
E0 = -0.50 V
14] Calculate the standard state cell potential for the following. 14
Cu(s)/Cu2+(aq)//K+(aq)/K(s)
15] What is the standard state reduction potential for the following reaction? 15
AgBr(s) + e- = Ag(s) + BrAg+ + e- = Ag(s)
E0 = 0.799 V
AgBr(s) = Ag+ + BrKsp = 5.0e-13
16] Calculate Ecell and K for this reaction. Pt/Cr3+(2.00e-4M), Cr2+(1.00e-3M)//Pb2+(6.50e-2M)/Pb. 16
17] What is E for the following if Cl- concentration is 0.50 M? 17
AgCl(s) + e- = Ag(s) + ClE0 = 0.222 V
18] Calculate E0 for 18
Ni(CN)4-(aq) + 2e- = Ni(s) + 4 CNGiven Ni2+ + 2e- = Ni(s)
E0 = -0.250 V  {Ni(CN)4-} = 1.0e22
19] What is the Ksp of AgCl given: 19
Ag+ + e- = Ag(s)
AgCl(s) + e- = Ag(s) + ClNote that 2.303RT/nF = 0.0592 V.
E0 = 0.799 V
E0 = 0.222 V
20] Calculate Ecell for Cd(s)/[CdCl2](aq) = 1.0 M//[AgNO3](aq) = 1.0 M/Ag(s) 20
a] 1.201 V
b] -1.201 V
c] 0.566 V
d] 0.997 V
e] -0.566 V
21] Calculate the standard potential for the following half-reaction, given the Ksp for Pd(OH)2 is
3.0 x 10-28 and the standard potential for Pd2+ + 2e- = Pd(s) is 0.915 V.21
Pd(OH)2(s) + 2e- = Pd(s) + 2OH
22] Explain what E0’ is and why is it preferred over E0 in biochemistry? 22
23] Calculate E0’ for H2C2O4 + 2H+ + 2e- = 2HCO2H
E0 = 0.204 V 23
Answers
1
b
To provide a stable potential in which the electrode reaction can be compared to 2H+(aq) + 2e= H2(g) E0 = 0.00 V
2
3
A3-
4
-0.762 V
5
E0cell = 0.0000 – (-2.7143) V
6
0.775 V
7
Cathode
8
E = -0.141 –(-0.44) = 0.30 V
9
E = -0.44 – (0.0592/2) log 1/Ksp = -0.756 V
10
Ecell = Ecath – Eanod = 2.890 – 0.771 = 2.119 V
11
E = 0.796 – 0.0592/2 log 1/[Hg22+]
Ksp = 7.4e-7 = [Hg22+][SO42-]
[Hg22+] = 7.4e-7/[SO42-]
E = 0.796 – 0.0592/2 log [SO42-]/7.4e-7 = 0.615 V
12
E = E0 – 0.0592 log 1/[H+] = 0.0000 – 0.0592 log 1/[1.00e-5] = -0.296 V
13
D+
14
Ecell = Ecath – Eanod = -2.936 – 0.339 = -3.275 V Use your book’s table of standard reduction
potentials.
15
E = E0(Ag+/Ag) – 0.0592 log 1/[Ag+]
Realize that
Ksp = [Ag+] [Br-]
[Ag+] = Ksp / [Br-]
sub into Nernst Eqn above
E = E0(Ag+/Ag) – 0.0592 log [Br-]/Ksplet [Br-] = 1 for standard state conditions
E0 = 0.799 – 0.0592 log 1/5.00e-13 = 0.0708 V
16
Pt/Cr3+(2.00e-4M), Cr2+(1.00e-3M)//Pb2+(6.50e-2M)/Pb
Ecath = -0.126 – (0.0592/2) log 1/6.5e-2 = -0.161 V
Eanod = -0.408 – 0.0592 log (1.00e-3/2.00e-4) = -0.449 V
Ecell = -0.161 –(-0.449) = 0.288 V
E0cell = -0.126 –(-0.408) = 0.282
G0 = -nFE0 = -2*96484*0.282 = - 5.44e4 J
G0 = -RT lnK
lnK = -G0 / RT =-(-5.44e4 J) / 8.314 * 298K
K = 3.43e9
17
E = E0 - 0.0592 log [Cl-] = 0.222 – 0.0592 log (0.50) = 0.240 V
18
E = -0.250 – (0.0592/2) log 1/[Ni2+]
 {Ni(CN)4-} = 1.0e22 = [Ni(CN)4-] / [Ni2+][CN-]4
[CN-]41.0e22 /[Ni(CN)4-] = 1/[Ni2+]
E = -0.250 – (0.0592/2) log {[CN-]41.0e22 /[Ni(CN)4-]}
E0 = -0.250 – (0.0592/2) log 1.0e22
E0 = -0.90 V
19
rxn: AgCl(s) = Ag+ + Cl-
add the following
Ag(s) = Ag+ + e-
E0 = 0.799 V
AgCl(s) + e- = Ag(s) + Cl-
E0 = 0.222 V
Ecell = 0.222 – 0.799 V = -0.577 V
∆G = -RT ln Ksp = -nFE
Ksp = 10^(-0.577/0.0592) = 1.79e-10
20
Anode: Cd = Cd2+ + 2e-
E0 = -0.402 V
Cathode Ag+ + e- = Ag(s)
E0 = 0.799 V
All at 1M concentration
Ecell = E0cell = Ecath - Eanod = 0.799 –(-0.402) V = 1.201 V
21
E = E0Pd2+/Pd – (0.0592 / 2) log 1 / [Pd2+]
Ksp = 3.0 x 10-28 = [Pd2+][OH-]2
[Pd2+] = 3.0 x 10-28 / [OH-]2
Standard potential conditions [OH-] = 1.0 M
[Pd2+] = 3.0 x 10-28
E = E0Pd2+/Pd – (0.0592 / 2) log 1 / 3.0 x 10-28
E = 0.915 V – (0.0592 / 2) log 1/ 3.0 x 10-28
E0Pd(OH-)2/Pd = -0.100 V
E0’ is the conditional from of E0 where it is assumed that the pH is 7 rather than 0. See
discussion in text
22
23
H2C2O4 is oxalic acid and HCO2H is formic acid.
E  E0 
0.0592
log Q
n
E  0.204 
[ HCO H ]2
0.0592
log  2 2
2
[ H ] [ H 2 C 2 O4 ]
#1
assume pH 7 and all else 1 M
First method is an approximation:
E 0 '  0.204 
0.0592
1
log 7 2  0.210V
2
[10 ]
We must consider that the [HCO2H] and the [H2C2O4] are involved in hydrolysis
reactions. The relative concentrations of each are described as:
 HCO H
2
[ HCO2 H ]
[H  ]


FHCO2 H
K a  [H  ]
[ H 2 C2 O4 ] 
[ HCO2 H ] 
[ H  ]FHCO2 H
K a  [H  ]
[ H  ]2 FH C O
2 2
4

K a1 K a 2  K a1[ H ]  [ H  ]2
#1 now becomes
2
E  0.204 
0.0592
log
2
 [ H  ]FHCO2 H 


 K a  [H  ] 


 2
[ H ] FH C O


2 2 4

[ H  ]2 
 K a1 K a 2  K a1 [ H  ]  [ H  ] 2 


[H+] = 1e7 exact, Ka = 1.80e-4, Ka1 = 5.60e-2, Ka2 = 5.42e-5, F = 1 M exact
E0’ = -0.268 V